Problem: Complete the square to solve for $x$. $4x^{2}+16x+15 = 0$
First, divide the polynomial by $4$ , the coefficient of the $x^2$ term. $x^2 + 4x + \dfrac{15}{4} = 0$ Move the constant term to the right side of the equation. $x^2 + 4x = -\dfrac{15}{4}$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $4$ , so half of it would be $2$ , and squaring it gives us ${4}$ $x^2 + 4x { + 4} = -\dfrac{15}{4} { + 4}$ We can now rewrite the left side of the equation as a squared term. $( x + 2 )^2 = \dfrac{1}{4}$ Take the square root of both sides. $x + 2 = \pm\dfrac{1}{2}$ Isolate $x$ to find the solution(s). $x = -2\pm\dfrac{1}{2}$ The solutions are: $x = -\dfrac{3}{2} \text{ or } x = -\dfrac{5}{2}$ We already found the completed square: $( x + 2 )^2 = \dfrac{1}{4}$